Let $$a, b \in R$$, $$(a \neq 0)$$. If the function $$f$$, defined as
$$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \leq x \lt 1 \\ a, & 1 \leq x \lt \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \leq x \lt 8 \end{cases}$$
is continuous in the interval $$[0, \infty)$$, then an ordered pair $$(a, b)$$ can be

We are told that the real-valued constants $$a$$ and $$b$$ make the piece-wise function

$$ f(x)= \begin{cases} \dfrac{2x^{2}}{a}, & 0\le x\lt 1\\[4pt] a, & 1\le x\lt \sqrt2\\[4pt] \dfrac{2b^{2}-4b}{x^{3}}, & \sqrt2\le x\lt 8 \end{cases} $$

continuous on the whole interval $$[0,\infty)$$. A function is continuous at a point when

$$\lim_{x\to c^{-}}f(x)=f(c)=\lim_{x\to c^{+}}f(x).$$

Inside every open sub-interval the defining formula itself is continuous, so the only possible trouble points are where the formula changes, namely at $$x=1$$ and $$x=\sqrt2$$ (the point $$x=0$$ is safe, because the single formula that covers it already gives $$f(0)=0$$ and the limit as $$x\to0^{+}$$ is also $$0$$).

Condition at $$x=1$$

The limit from the left uses the first formula:

$$ \lim_{x\to1^{-}}\dfrac{2x^{2}}{a}=\dfrac{2(1)^{2}}{a}=\dfrac2a. $$

The value (and the limit from the right) at $$x=1$$ comes from the second formula:

$$ f(1)=a. $$

For continuity we require

$$ \dfrac2a = a. $$

Multiplying both sides by $$a$$ gives

$$ 2 = a^{2}\quad\Longrightarrow\quad a=\pm\sqrt2. $$

Thus only two values of $$a$$ are possible:

$$ a = \sqrt2 \quad\text{or}\quad a=-\sqrt2. $$

Condition at $$x=\sqrt2$$

The limit from the left (where $$x\lt \sqrt2$$) is simply the constant value $$a$$:

$$ \lim_{x\to(\sqrt2)^{-}}f(x)=a. $$

The value (and the limit from the right) at $$x=\sqrt2$$ is obtained from the third formula:

$$ f(\sqrt2)=\frac{\,2b^{2}-4b\,}{(\sqrt2)^{3}}. $$

Because $$(\sqrt2)^{3}=\sqrt2\cdot2=2\sqrt2$$, we have

$$ f(\sqrt2)=\frac{2b^{2}-4b}{2\sqrt2} =\frac{2b(b-2)}{2\sqrt2} =\frac{b(b-2)}{\sqrt2}. $$

Continuity at $$x=\sqrt2$$ therefore demands

$$ a=\frac{b(b-2)}{\sqrt2}. \cdots(1) $$

Substituting the possible values of $$a$$

Case 1: $$a=\sqrt2$$. Then equation (1) becomes

$$ \sqrt2=\frac{b(b-2)}{\sqrt2}\quad\Longrightarrow\quad b(b-2)=2. \cdots(2) $$

Case 2: $$a=-\sqrt2$$. Then equation (1) becomes

$$ -\sqrt2=\frac{b(b-2)}{\sqrt2}\quad\Longrightarrow\quad b(b-2)=-2. \cdots(3) $$

Checking the four given options

We now verify which listed ordered pair $$(a,b)$$ satisfies the appropriate relation.

• Option A: $$a=-\sqrt2,\; b=1-\sqrt3.$$
For this $$b(b-2)=(1-\sqrt3)(-1-\sqrt3)=2,$$ so we get equation (2), not (3). Hence Option A fails.

• Option B: $$a=\sqrt2,\; b=-1+\sqrt3.$$
Compute $$b(b-2)=(-1+\sqrt3)(-3+\sqrt3)=6-4\sqrt3\neq2,$$ so equation (2) is not satisfied. Option B fails.

• Option C: $$a=\sqrt2,\; b=1-\sqrt3.$$
Again $$b(b-2)=2,$$ which exactly fulfils equation (2). Therefore Option C works.

• Option D: $$a=-\sqrt2,\; b=1+\sqrt3.$$
Here $$b(b-2)=(1+\sqrt3)(-1+\sqrt3)=2,$$ but equation (3) needs $$b(b-2)=-2$$. Hence Option D fails.

Only Option C satisfies all the continuity conditions.

Hence, the correct answer is Option C.